on his 70th birthday
VECTOR SEMINORMS, SPACES WITH VECTOR NORM, AND REGULAR OPERATORS
ROMULUS CRISTESCU
The rst section of this paper deals with the properties of some type of vector seminorms. In the second section, vector spaces with vector norm are considered while in the third section some theorems about the regular operators are given.
Operators dened on a vector space with vector norm are considered in the fourth section. The terminology is that of [3].
AMS 2000 Subject Classication: 47860.
Key words: vector seminorm, regular operators.
1. VECTOR SEMINORMS
IfX is a vector space overR andY an ordered vector space, a mapping P ofX intoY is said to be a vector seminorm if
P(x1+x2)≤P(x1) +P(x2), ∀x1, x2 ∈X, P(αx) =|α|P(x), ∀α∈R, ∀x∈X.
IfX and Y are ordered vector spaces, a vector seminorm P :X →Y is said to be monotone if
0≤a≤b∈X⇒P(a)≤P(b) while it is said to be absolute monotone if
±x≤c∈X⇒P(x)≤P(c).
An ordered vector space is said to be (o)-complete if every majorized subset has a supremum.
The following theorem generalize some results given in [7] for real semi- norms.
Theorem 1.1. Let X be a directed vector space, G a majorized vector subspace of X, and Y an (o)-complete ordered vector space. If P0 :G→Y is
REV. ROUMAINE MATH. PURES APPL., 53 (2008), 56, 407418
a vector seminorm, then by putting
(1.1) P(x) = inf{P0(a) +P0(z) |a, z∈G; ±(x−z)≤a}, ∀x∈X, the following statements hold.
(i) The operator P : X → Y given by (1.1) is an absolute monotone vector seminorm and P(z)≤P0(z),∀z∈G.
(ii) If Q : X → Y is an absolute monotone vector seminorm and if Q(z)≤P0(z),∀z∈G, then Q(x)≤P(x),∀x∈X.
(iii) We haveP |G=P0if and only if the vector seminorm P0is absolute monotone.
Proof. (i) For any x ∈ X there exists the elementP(x) given by (1.1).
Indeed, since G is a majorizing vector subspace of the directed vector space X, for any elementsx∈X andz∈Gthere exists an elementa∈Gsuch that
±(x−z)≤a.On the other hand, we haveP0(c)≥0, ∀c∈G, and the space Y is(o)-complete.
As in the case of real seminorms, it is easily veried that the operatorP given by (1.1) is a vector seminorm.
We also have
(1.2) P(x) = inf{P0(b)|x≤b∈G}, ∀x∈X+.
Indeed, denoting by P0(x) the right-hand side in (1.2), we obviously have P(x) ≤ P0(x). Let now x ∈ X+ and ±(x−z) ≤ a with a, z ∈ G. Then 0≤x≤a+z∈G, whence
P0(x)≤P0(a+z)≤P0(a) +P0(z).
So, P0(x)≤P(x). Therefore (1.2) holds.
The vector seminorm P is absolute monotone. Indeed, if ± x≤v inX then, takinga∈Gsuch thatv≤a, it follows from±x≤athatP(x)≤P0(a) by (1.1). Then, by (1.2), we get P(x)≤P(v).
ObviouslyP(z)≤P0(z), ∀z∈G.
(ii) LetQ:X →Y be an absolute monotone vector seminorm such that Q(z)≤P0(z), ∀z∈G. Ifx∈X and±(x−z)≤awitha, z∈G, then
Q(x)≤Q(x−z) +Q(z)≤Q(a) +Q(z)≤P0(a) +P0(z) whenceQ(x)≤P(x).
(iii) Suppose that P0 is absolute monotone and let x ∈ G. Let also, z, a ∈ G such that ±(x−z) ≤a. We have P0(x−z) ≤ P0(a) whence, as in the preceding case regarding the vector seminormQ, we obtainP0(x)≤P(x). We also have P(x)≤P0(x)by statement (i).
Conversely, ifP |G=P0 thenP0 is obviously absolute monotone.
Denition 1.1. If X is a directed vector space and Y an (o)-complete ordered vector space, a vector seminorm P :X→Y is said to be solid if
P(x) = inf{P(a)| ± x≤a∈X}, ∀x∈X.
Remark 1.1. Let X be a directed vector space and Y an (o)-complete ordered vector space. IfGis a majorizing vector subspace ofXandP0 :G→Y a solid vector seminorm, then the formula
P(x) = inf{P0(a)| ±x≤a∈G}, ∀x∈X, yields a solid vector seminorm P :X →Y such thatP |G=P0.
Remark 1.2. Let X be a vector lattice and Y an (o)-complete ordered vector space. A vector seminorm P :X→ Y is solid if and only if |x1| ≤ |x2| inX impliesP(x1)≤P(x2). This assertion is veried as in the caseY =R.
Remark 1.3. LetX be a vector lattice,Y an(o)-complete ordered vector space and P : X → Y a solid vector seminorm. Then P is additive on the positive cone X+, if and only ifP is of the form
(1.3) P(x) =U(|x|), ∀x∈X,
whereU :X→Y is a positive linear operator. Indeed, ifP is additive on X+ then, by putting,
U(x) =P(x+)−P(x−), ∀x∈X,
we obtain a positive linear operator U :X → Y for which (1.3) holds. Con- versely, if U :X →Y is a positive linear operator, then by (1.3) we obviously obtain a solid vector seminorm.
We recall now that ifXandY are vector lattices, an operatorU :X→Y is said [3] to be disjunctive if x1 ⊥x2 inX impliesU(x1)⊥U(x2).
Theorem 1.2. If X and Y are vector lattices and P : X → Y is a monotone and disjunctive vector seminorm, then
(1.4) P(x1∧x2) =P(x1)∧P(x2), ∀x1, x2 ∈X+, and if x1∧x2 =0 in X, then
(1.5) P(x1∨x2) =P(x1)∨P(x2).
Proof. Letx1, x2 ∈X+. We have
(x1−(x1∧x2))∧(x2−(x1∧x2)) =0, whence
(1.6) P(x1−(x1∧x2))∧P(x2−(x1∧x2)) =0.
On the other hand
0≤P(xi)−P(x1∧x2)≤P(xi−(x1∧x2)), i= 1,2, and, by (1.6),
(P(x1)−P(x1∧x2))∧(P(x2)−P(x1∧x2)) =0.
Therefore, (1.4) holds.
Ifx1∧x2 =0 thenx1∨x2=x1+x2, hence
P(x1∨x2) =P(x1+x2)≤P(x1) +P(x2) =P(x1)∨P(x2) since P(x1)⊥P(x2). On the other hand, we have
P(x1)∨P(x2)≤P(x1∨x2), so that (1.5) holds.
Remark 1.4. Let X and Y be vector lattices and let P :X → Y be a solid vector seminorm. Then P is disjunctive if and only if x1∧x2 =0 inX implies P(x1)∧P(x2) =0.
Theorem 1.3. Let X be a vector lattice and Y an (o)-complete ordered vector space. If P :X →Y is a (non-zero) solid vector seminorm, then there exists a (non-zero) positive linear operator U :X→Y such that
(1.7) |U(x)| ≤P(x), ∀x∈X.
Proof. The proof is similar to that given in the case of functionals [4].
By putting
Q(x) =P(x+), ∀x∈X,
we obtain a sublinear operator Q:X →Y. Let us consider an elementa >0 of X such that P(a) > 0. By a corollary of the Hahn-Banach-Kantorovich theorem [5] there exists a linear operator U :X → Y such thatU(a) = Q(a) and U(x) ≤ Q(x), ∀x ∈ X. It is easy to see that U is positive and that inequality (1.7) holds.
Theorem 1.4. LetXbe a vector lattice,Y an(o)-complete ordered vector space and P(X, Y) the set of all solid vector seminorms mapping X into Y. If Pi ∈ P(X, Y), i= 1,2, then, with respect to the pointwise order in the set P(X, Y),there exists P1∧P2 in this set, and we have
(P1∧P2)(x) = inf{P1(a) +P2(b)|a, b∈X+; a+b=|x|}, ∀x∈X.
The proof is similar to that given in the caseY =R(see [2]).
2. SPACES WITH VECTOR NORM
LetF be a real vector space and X a vector lattice. A vector seminorm P :F →X is said to be a vector norm ifP(x) =0impliesx=0. If the vector spaceFis endowed with a vector normP :F →X, thenF is called av-normed vector space and is denoted by FX. We shall also writekfkX =P(f).
LetFX be a v-normed vector space.
A subset A of FX is said to be (v)-bounded if there existsx0 ∈X such that kfkX ≤x0, ∀f ∈A.
A sequence (fn)n∈N of elements of the space FX is said to be a (v)- Cauchy sequence if there exists a sequence (an)n∈Nof elements ofX such that an ↓
n∈N
0 and
(2.1) kfi−fjkX ≤an, ∀i, j≥n.
We say that a sequence (fn)n∈N (v)-converges to an element f ∈ FX if the sequence (kfn−fkX)n∈
N (o)-converges (i.e. converges with respect to the order) to 0 in the space X. The element f is called the (v)-limit of the sequence (fn)n∈N, and we writef = (v)-lim
n fn.
The space FX is said to be (v)-complete if any (v)-Cauchy sequence of elements of FX(v)-converges to an element of the space.
The space FX is called a strictly v-normed vector space if whenever kfkX = a1 +a2 with ai ∈ X+, i = 1,2, there exists fi ∈ FX, i = 1,2, such thatkfikX =ai, i= 1,2, and f =f1+f2.
A strictlyv-normed vector space which is also (v)-complete is said to be a space of type (BK).
The notion of a(v)-convergent series in a v-normed vector space can be introduced in a natural manner:
(v)-
∞
X
n=1
fn= (v)-lim
n n
X
i=1
fi
if the right-hand member (v)-limit exists.
IfFX is av-normed vector space with respect to an Archimedean vector lattice X, a sequence(fn)n∈N of elements ofFX is said to(vρ)-converge to an element f ∈FX if the sequence (kfn−fkX)n∈
N (ρ)-converges (i.e. converges with regulator [3]) to 0 in the spaceX.
Remark 2.1. If a(v)-Cauchy sequence(fn)n∈Nof elements of av-normed vector space has a (v)-convergent subsequence, then the sequence (fn)n∈N is (v)-convergent.
In the next theorem, by regular space we mean (as in [4]) an Archimedean vector lattice in which any (o)-convergent sequence is also(ρ)-convergent.
Theorem 2.1. Let FX be a v-normed vector space and consider the following two conditions:
(i) the space FX is (v)-complete;
(ii) if a series of the form P∞
n=1
kfnkX, fn ∈ FX, is (o)-convergent, then the series P∞
n=1
fn is (v)-convergent.
Condition (i) implies (ii), and if X is a σ-complete vector lattice which is also a regular space, then conditions (i) and (ii) are equivalent.
Proof. Suppose that condition (i) is satised and let(fn)n∈Nbe a sequence of elements of FX such that the series P
n
kfnkX be(o)-convergent. Denoting
sn=
n
X
i=1
fi, n∈N,
there exists a sequence (wn)n∈N of elements ofX such thatwn ↓
n∈N
0 and ksn−smkX ≤wm, m, n∈N; m≤n.
Therefore,(sn)n∈Nis a(v)-Cauchy sequence and sinceFX is(v)-complete, the sequence (sn)n∈N is(v)-convergent. Consequently, (i) implies (ii).
Suppose now thatX is aσ-complete vector lattice which also is a regular space and suppose that condition (ii) is satised. Let (fn)n∈Nbe a(v)-Cauchy sequence of elements of FX.Let (an)n∈N be a sequence of elements ofX such that an ↓
n∈N
0 and
(2.1) kfi−fjkX ≤an, i, j≥n.
SinceX is a regular space, there exists w∈X+ and a sequence(εn)n∈N
of real numbers, such thatεn ↓
n∈N
0andan≤εnw,∀n∈N. According to (2.1) there exists a strictly increasing sequence(jn)n∈Nof natural numbers such that
(2.2)
fjn+1−fjn X ≤ 1
2nw, ∀n∈N.
Since the vector lattice X is σ-complete, it follows from (2.2) that the series corresponding to the sequence (kfjn+1 −fjnkX)n∈N is (o)-convergent.
By hypothesis, the series P
n
(fjn+1 −fjn) is (v)-convergent. It follows from the equation
fjn+1 =fj1 +
n
X
k=1
fjk+1−fjk
, ∀n∈N,
that the sequence (fjn)n∈N is (v)-convergent while, by Remark 2.1, the se- quence (fn)n∈N is (v)-convergent. Consequently, if X is a σ-complete vector lattice which also is a regular space, then conditions (i) and (ii) are equiva- lent.
Theorem 2.2. Let X and Y be complete vector lattices and Ga vector sublattice of X endowed with a solid vector norm with values in Y. If X is a Dedekind extension of G, then the vector norm given on Gcan be extended to a solid vector norm on X by the formula
(2.3) kxkY = inf{kakY | |x| ≤a∈G}, ∀x∈X.
Proof. Ifx1, x2 ∈X then, denoting
A={kakY |a=b+c, |x1| ≤b∈G, |x2| ≤c∈G}, B ={kbkY | |x1| ≤b∈G}, C ={kckY | |x2| ≤c∈G}, we have
kx1+x2kY ≤infA≤inf (B+C) = infB+ infC =kx1kY +kx2kY . On the other hand, one can easily verify that
kαxkY =|α| kxkY , ∀α∈R, ∀x∈X.
If06=x∈Xthen there existsb∈Gsuch that0< b≤ |x|. If|x| ≤a∈G then 0< b≤a, hence
0<kbkY ≤ kakY .
It follows is immediately that kxkY 6= 0. Therefore, by (2.3), we obtain a vector norm on X. It is easily veried that this vector norm is solid and that it is an extension of the vector norm given on G.
3. REGULAR OPERATORS
LetXbe a vector lattice andY a complete vector lattice. We shall denote by R(X, Y) the set of all regular operators which mapX intoY. It is known thatR(X, Y) is a complete vector lattice with respect to the usual operations and the order given by the cone of positive operators. We shall also denote Xr =R(X,R).
As in the rst section (see Theorem 1.4) we denote byP(X, Y)the set of all solid vector seminorms onX intoY and shall consider the pointwise order in this set.
Theorem 3.1. Let Ui ∈ R(X, Y)and Pi ∈ P(X, Y),i= 1,2,such that (3.1) |Ui(x)| ≤Pi(x), ∀x∈X, i= 1,2.
Then the inequality
(3.2) |(|U1| ∧ |U2|) (x)| ≤(P1∧P2) (x), ∀x∈X.
holds.
Proof. It follows from (3.1) that
| |Ui|(x)| ≤Pi(x), ∀x∈X, i= 1,2, and by Theorem 1.4 we have
|(|U1| ∧ |U2|) (x)| ≤(|U1| ∧ |U2|) (|x|) =
= inf{|U1|(a) +|U2|(b)|a, b∈X+; a+b=|x|} ≤
≤inf{P1(a) +P2(b)|a, b∈X+; a+b=|x|}= (P1∧P2) (x).
Therefore, (3.2) holds.
Denition 3.1. Letp:X →Rbe a seminorm. An operatorU :X →Y is said to bep-bounded if there existsy0 ∈Y such that|U(x)| ≤p(x)y0, ∀x∈X.
Remark 3.1. Theorem 3.1 implies the following proposition given in [2]. If U1, U2 ∈ R(X, Y),p1, p2 are solid real seminorms on X andUi ispi-bounded, i= 1,2, then|U1| ∧ |U2|is p1∧p2-bounded.
Theorem 3.2. Let P :X→Y be a solid vector seminorm, Ga normal subspace of X and V0 :G→Y a positive linear operator such that
(3.3) |V0(a)| ≤P(a), ∀a∈G.
Then there exists a positive linear operator V :X →Y such that V |G=V0
and
(3.4) |V(x)| ≤P(x), ∀x∈X.
Proof. By the Hahn-Banach-Kantorovich theorem [5], there exists a linear operator U : X → Y such that U | G = V0 and U(x) ≤ P(x), ∀x ∈ X. It follows that U is an (o)-bounded linear operator, hence U ∈ R(X, Y). Let us now consider the positive part U+ of U. If x ∈ X+ then it follows from 0≤z≤x that
U(z)≤P(z)≤P(x), hence
U+(x) = sup{U(z)|0≤z≤x} ≤P(x).
For any x∈X we now have
|U+(x)| ≤U+(|x|)≤P(|x|) =P(x).
Therefore, putting V =U+, inequality (3.4) holds. On the other hand, since G is a normal subspace ofX, if 0≤a∈Gthen[0, a]⊂G. It follows that
V(a) = supU([0, a]) =V0(a)
and, Gbeing a vector sublattice ofX, we haveV(x) =V0(x),∀x∈G.
Remark 3.2. It follows by Theorems 3.2 and 3.1 that if G is a normal subspace of the spaceX, if 0≤Vi ∈ R(G, Y), Pi∈ P(X, Y), i= 1,2, and
|Vi(a)| ≤Pi(a), ∀a∈G, i= 1,2,
then there exists a positive V ∈ R(X, Y)such thatV |G=V1∧V2 and
|V(x)| ≤(P1∧P2) (x), ∀x∈X.
4. LINEAR OPERATORS ON v-NORMED VECTOR SPACES
LetX, Y be vector lattices andFX, GY v-normed vector spaces. We recall that an operator U :FX → GY is said to be a(v)-regular operator [3] if it is additive and if there exists a positive linear operatorV :X →Y such that (4.1) kU(f)kY ≤V(kfkX), ∀f ∈FX.
We shall denote byRv(FX, GY)the set of all(v)-regular operators which map FX intoGY.
If the spaceY is Archimedean, then a(v)-regular operator which maps FX intoGY is a linear operator.
Remark 4.1. If Y is a complete vector lattice and in the spaces X, Y we consider the modulus as vector norm, then Rv(X, Y) =R(X, Y).
Remark 4.2. By a proposition given in [3], if FX is a strictly v-normed vector space, Y is a complete vector lattice and U ∈ Rv(FX, GY), then there exists the smallest positive linear operator V which satises inequal- ity (4.1). In this case we shall denote such an operator by kUkR. The map P : Rv(FX, GY) → R(X, Y) given by the formula P(U) = kUkR is a vector norm on the vector space Rv(FX, GY).
We recall now that a subset A of an Archimedean vector lattice is said to be (o)-annihilating [3] if for any sequence (xn)n∈N of elements of A and any sequence (αn)n∈N of real numbers which converges to zero, one has (o)- limn αnxn= 0. An Archimedean vector latticeZis said to be of(o)-boundedness type if any (o)-annihilating subset ofZ is(o)-bounded.
The following theorem generalizes a result given in [8].
Theorem 4.1. Let X, Y be Archimedean vector lattices and FX, GY v- normed vector spaces. Let (Un)n∈Nbe a sequence of linear operators which map FX into GY.Consider the following statements.
(i) For any(v)-bounded subsetAof FX, the set S
n∈N
Un(A)is (v)-bounded.
(ii) If (vρ)-lim
n fn=0inFX, then(vρ)-lim
n Ujn(fn) =0for any sequence (jn)n∈Nof natural numbers.
Statement (i) implies (ii) and if Y is of (o)-boundedness type, then (i) and (ii) are equivalent.
Proof. Suppose (i) holds and let (vρ)-lim
n fn =0 in FX. Then [5] there exists a sequence (λn)n∈N of natural numbers such that λn ↑
n∈N
∞ and (ρ)- limn λnkfnkX =0. It follows that the sequence (λnfn)n∈N is(v)-bounded. By hypothesis, there exists y0∈Y such that
(4.2) kUk(λnfn)kY ≤y0, ∀k, n∈N.
If (jn)n∈N is an arbitrary sequence of natural numbers, then by (4.2) we have kUjn(fn)kY ≤ 1
λny0, ∀n∈N, hence (vρ)-lim
n Ujnfn=0. Consequently, (i)⇒ (ii).
Let us now assume that Y is of (o)-boundedness type and suppose (ii) holds. Let Abe a (v)-bounded subset of the space FX. Denote
B = [
k∈N
Uk(A)
and let us consider a sequence (gn)n∈N of elements of B. Let jn ∈ N and fn ∈ A with gn = Ujn(fn). Consider now a sequence (αn)n∈N of real num- bers convergent to zero. Since the sequence (kfnkX)n∈
N is (o)-bounded, the sequence (αnkfnkX)n∈N is ρ-convergent to 0, hence (vρ)-lim
n αnfn = 0. By hypothesis,
(vρ)-lim
n Ujn(αnfn) =0 hence (ρ)-lim
n αnkgnkY = 0. Consequently, the set {kgkY | g ∈ B} is (o)- annihilating, therefore it is (o)-bounded. In other words, the set B is (v)- bounded.
Remark 4.3. LetFX be a strictlyv-normed vector space,GY av-normed vector space with Y complete vector lattice and Un ∈ Rv(FX, GY), n ∈ N.
If the sequence (kUnkR)n∈
N is (o)-bounded inR(X, Y) then statement (i) of Theorem 4.1 holds.
In the next theorem we consider the modulus as vector norm in a complete vector lattice Y.
Theorem 4.2. Let F be a vector lattice and E a vector sublattice of F. Consider a solid vector norm on F with values in a vector lattice X such that FX andEX are strictlyv-normed vector spaces. If Y is a complete vector lattice and U0 :EX →Y is a positive (v)-regular operator, then there exists a positive (v)-regular operator U :FX →Y such that U |E =U0 and kUkR=kU0kR.
Proof. Denoting byF+ the positive cone of the spaceF and putting Q(f) = inf
kU0kR(khkX)|f ≤h∈F+ , ∀f ∈F, we obtain a sublinear operator Q:F →Y. It is easily veried that (4.3) Q(f)≤ kU0kR(kfkX), ∀f ∈F.
The inequality
(4.4) U0(f)≤Q(f), ∀f ∈E,
also holds. Indeed, if f ∈E and f ≤h ∈F+, then f ≤f+ ≤h withf+ ∈E and we have
U0(f)≤U0(f+)≤ kU0kR(kf+kX)≤ kU0kR(khkX) whence (4.4) follows.
By the Hahn-Banach-Kantorovich theorem [5], there exists a linear ope- rator U :F →X such thatU |E=U0 and
(4.5) U(f)≤Q(f), ∀f ∈F,
whence, by (4.3), |U(f)| ≤ kU0kR(kfkX), ∀f ∈F.
Therefore, U ∈ Rv(FX, Y) and kUkR = kU0kR. On the other hand, if f ∈F and f ≤0, then Q(f) =0 and it follows by (4.5) thatU(f) ≤0, that is, U is a positive operator.
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Received 16 July 2008 University of Bucharest
Faculty of Mathematics and Computer Science Str. Academiei 14
010014 Bucharest, Romania
